The trick is to pass the name of the variable, not its expanded value.
Use
func out
instead of
func $out
In the function, set the value of the out parameter as
eval "$1=expr"
In the function, copy the value of the parameter to a local variable with
eval "L_VAR=\${$1}"
When using local variables, it is important that the name of the variable from the caller is not the same name as the local variable in the function. The local in the function is deleted when the function ends.
#!/bin/bash #$1 is name of out param, $2 is name of out param func () { local loc=8 local diff=9 local same=10 eval "$1=$diff" eval "$2=$same" } loc=7 func out same echo "loc = $loc" #loc = 7 echo "out = $out" #out = 9 echo "same = $same" #same =
